(a) Solve : \((x – 2)(x – 3) = 12\).
(b) 
In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
Explanation
(a) \((x - 2)(x - 3) = 12\)
\(x^{2} - 3x - 2x + 6 = 12 \implies x^{2} - 5x + 6 - 12 = 0\)
\(x^{2} - 5x - 6 = 0\)
\(x^{2} - 6x + x - 6 = 0 \implies x(x - 6) + 1(x - 6) = 0\)
\( (x - 6)(x + 1) = 0 \implies \text{x = 6 or -1}\)
(b) 
The shaded portion comprises two shaded segments labelled A and B.
Area of segment A = \(\frac{60°}{360°} \times \pi r^{2} - \frac{1}{2} r^{2} \sin 60°\)
\(\frac{1}{6} \times \frac{22}{7} \times 7 \times 7 - \frac{1}{2} \times 7 \times 7 \times 0.8660\)
= \(25.667 - 21.217\)
= \(4.45 cm^{2}\)
Also, the area of the segment B = \(4.45 cm^{2}\).
Hence, the area of the shaded portion = \(4.45 cm^{2} + 4.45 cm^{2}\)
= \(8.9 cm^{2}\)
\(\approxeq 9 cm^{2}\) (to the nearest whole number).